# How do you differentiate f(x)=x(1+e^(x^2))^(1/5) using the chain rule?

Dec 14, 2015

$f ' \left(x\right) = \frac{5 + 5 {e}^{{x}^{2}} + 2 {x}^{2} {e}^{{x}^{2}}}{5 {\left(1 + {e}^{{x}^{2}}\right)}^{\frac{4}{5}}}$

#### Explanation:

First, use the product rule:

f'(x)=(1+e^(x^2))^(1/5)color(blue)(d/dx[x])+xcolor(green)(d/dx[(1+e^(x^2))^(1/5)]

Find each derivative separately.

color(blue)(d/dx[x]=1

The following requires heavy use of the chain rule:

color(green)(d/dx[(1+e^(x^2))^(1/5)])=1/5(1+e^(x^2))^(-4/5)color(red)(d/dx[1+e^(x^2)]

color(red)(d/dx[1+e^(x^2)])=e^(x^2)d/dx[x^2]=color(red)(2xe^(x^2)

Plug this back in.

color(green)(d/dx[(1+e^(x^2))^(1/5)])=(2xe^(x^2)(1+e^(x^2))^(-4/5))/5=color(green)((2xe^(x^2))/(5(1+e^(x^2))^(4/5))

Finally, plug these in to find $f ' \left(x\right)$.

$f ' \left(x\right) = {\left(1 + {e}^{{x}^{2}}\right)}^{\frac{1}{5}} + \frac{2 {x}^{2} {e}^{{x}^{2}}}{5 {\left(1 + {e}^{{x}^{2}}\right)}^{\frac{4}{5}}}$

Multiply the first term for a common denominator.

$f ' \left(x\right) = \frac{5 + 5 {e}^{{x}^{2}} + 2 {x}^{2} {e}^{{x}^{2}}}{5 {\left(1 + {e}^{{x}^{2}}\right)}^{\frac{4}{5}}}$